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Kitty [74]
2 years ago
10

Demonstrate an understanding of Stoichiometry by describing and calculating the components required for a stoichiometric evaluat

ion and performing stoichiometric evaluations for determining the limiting reactant and percent yield.
Chemistry
1 answer:
NeTakaya2 years ago
8 0

Explanation:

Iam sorry I don't know but why Iam messaging iss because when more people message it usually appears to more people so someone else will be able to help you:)

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Match the vocabulary with the definitions.
Fiesta28 [93]

1: Decomposition reaction
2: Combination reaction
3: product
4: Reactant
8 0
3 years ago
True or false
sweet-ann [11.9K]

Answer:

True.

But it only changes in physical change.

How?

Explanation:

The chemical reaction produces a new substance with new and different physical and chemical properties. Matter is never destroyed or created in chemical reactions. The particles of one substance are rearranged to form a new substance.

In a physical change, a substance's physical properties may change.

A chemical change is a permanent change. A Physical change affects only physical properties i.e. shape, size, etc. ... Some examples of physical change are freezing of water, melting of wax, boiling of water, etc. A few examples of chemical change are digestion of food, burning of coal, rusting, etc.

Hope this helps!

8 0
3 years ago
And I living organism what is a tissue
Leviafan [203]
It is an ensemble of similar cells
4 0
3 years ago
Read 2 more answers
Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit
Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

0.0811 M=\frac{n}{0.0017 L}

n = 0.0001378 mol

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0
3 years ago
How many molecules Of H20 are<br> equivalent to 97.2 g H20?<br> (H = 1.008 g/mol, O = 16.00 g/mol)
AleksandrR [38]

Answer:

m(H₂O) = 97,2 g.n(H₂O) = m(H₂O) ÷ M(H₂O).n(H₂O) = 97,2 g ÷ 18

Explanation:

8 0
2 years ago
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